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A certain musical scale has has 13 notes, each having a

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A certain musical scale has has 13 notes, each having a [#permalink] New post 01 Jan 2004, 22:34
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A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Feb 2013, 05:01, edited 1 time in total.
Edited the question and added OA.
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Re: OG PS #434-- musical scale [#permalink] New post 02 Jan 2004, 01:49
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stoolfi wrote:

On another thread, someone with a very good GMAT score said he couldn't really understand this question. It is posted for your perusal:

A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)



let the constant be k

F1 = 440

F2 = 440k

F3 = 440 k * k = 440 * k^2

F13= 440 * k^12

we know F13 = 2 *F1 = 2 * 440 = 880

880/440 = k^12

k = twelfth root of 2

for F7...

F7 = 440 * k^6 ( as we wrote for F2 and F3)
F7 = 440 * (twelfth root of 2) ^ 6

F7 = 440 * sqrt (2)


Answer A
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 [#permalink] New post 02 Jan 2004, 05:01
I agree Stoolfi, one of those wacky questions that involve figuring out the hidden "trick". The GMAT and its bag of tricks! :magic
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Re: OG PS #434-- musical scale [#permalink] New post 09 Nov 2011, 03:00
Praetorian wrote:
stoolfi wrote:

On another thread, someone with a very good GMAT score said he couldn't really understand this question. It is posted for your perusal:

A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)



let the constant be k

F1 = 440

F2 = 440k

F3 = 440 k * k = 440 * k^2

F13= 440 * k^12

we know F13 = 2 *F1 = 2 * 440 = 880

880/440 = k^12

k = twelfth root of 2

for F7...

F7 = 440 * k^6 ( as we wrote for F2 and F3)
F7 = 440 * (twelfth root of 2) ^ 6

F7 = 440 * sqrt (2)


Answer A


The question stem says that : the ratio of frequency to next higer frequency is fixed constant ...

Doesnt that mean F1/F2 = k

F2/F3 = k^2 and something like that .....???and not F2/F1 = k which is F2 = kF1???
:roll: :roll: :roll: :roll: :roll: :roll: :roll:
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Re: On another thread, someone with a very good GMAT score said [#permalink] New post 09 Nov 2011, 21:27
Siddhans,

It says the ratio of a frequency to the next higher frequency is a constant. f2/f1=f3/f2=f4/f3=.....=fixed number. This is an example of a geometric progression.

Here is a video explanation:

http://www.gmatquantum.com/og10-journal ... ition.html

Dabral

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Re: OG PS #434-- musical scale [#permalink] New post 22 Feb 2013, 03:38
Praetorian wrote:
stoolfi wrote:

On another thread, someone with a very good GMAT score said he couldn't really understand this question. It is posted for your perusal:

A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)



let the constant be k

F1 = 440

F2 = 440k

F3 = 440 k * k = 440 * k^2

F13= 440 * k^12

we know F13 = 2 *F1 = 2 * 440 = 880

880/440 = k^12

k = twelfth root of 2

for F7...

F7 = 440 * k^6 ( as we wrote for F2 and F3)
F7 = 440 * (twelfth root of 2) ^ 6

F7 = 440 * sqrt (2)


Answer A


Why do you multiply and not add?

Like
1. 440
2. 440 + k and so on?
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Re: A certain musical scale has has 13 notes, each having a [#permalink] New post 22 Feb 2013, 04:48
Lowest frequency = 440
Highest frequency = 880
Lowest frequency (n) ^12 = Highest frequency
N^12 = 2 ---------------- 1
7th note = Lowest Frequency x (n)^6
7th note = 440 x (2)^6/12
Hence the answer is A

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Re: OG PS #434-- musical scale [#permalink] New post 22 Feb 2013, 05:12
Expert's post
karmapatell wrote:
Praetorian wrote:
stoolfi wrote:

On another thread, someone with a very good GMAT score said he couldn't really understand this question. It is posted for your perusal:

A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)



let the constant be k

F1 = 440

F2 = 440k

F3 = 440 k * k = 440 * k^2

F13= 440 * k^12

we know F13 = 2 *F1 = 2 * 440 = 880

880/440 = k^12

k = twelfth root of 2

for F7...

F7 = 440 * k^6 ( as we wrote for F2 and F3)
F7 = 440 * (twelfth root of 2) ^ 6

F7 = 440 * sqrt (2)


Answer A


Why do you multiply and not add?

Like
1. 440
2. 440 + k and so on?


Because we are given that the ratio of a frequency to the next higher frequency is a fixed constant: F2/F1=k.

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Re: OG PS #434-- musical scale [#permalink] New post 22 Feb 2013, 09:29
karmapatell wrote:
Praetorian wrote:
stoolfi wrote:

On another thread, someone with a very good GMAT score said he couldn't really understand this question. It is posted for your perusal:

A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)



let the constant be k

F1 = 440

F2 = 440k

F3 = 440 k * k = 440 * k^2

F13= 440 * k^12

we know F13 = 2 *F1 = 2 * 440 = 880

880/440 = k^12

k = twelfth root of 2

for F7...

F7 = 440 * k^6 ( as we wrote for F2 and F3)
F7 = 440 * (twelfth root of 2) ^ 6

F7 = 440 * sqrt (2)


Answer A


Why do you multiply and not add?

Like
1. 440
2. 440 + k and so on?




"For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant"

The question says that the ratio of two consecutive frequencies is constant. Hence we multiply.

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Re: A certain musical scale has has 13 notes, each having a [#permalink] New post 22 Feb 2013, 12:59
Shoudln't "the ratio of a frequency to the next higher frequency is a fixed constant" be interpreted as f1/f2 = k instead of f2/f1=k?

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Re: A certain musical scale has has 13 notes, each having a [#permalink] New post 22 Feb 2013, 21:51
Let's take a simple example

Series : 2 4 6 8
Now going forward the multiplying factor is 2 and backward the factor is 1/2

Hence if you go backwards in the question the factor will be 1/2 and still you will get the same answer.

Posted from my mobile device Image

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Re: A certain musical scale has has 13 notes, each having a [#permalink] New post 23 Jun 2013, 14:04
Since the ratio is constant it is a geometric progression

7th term is the middle term of the series

mean of the geometric series given by sqrt(a*b)

a= first term of the series = 440
b= last term of the series = 2*440

Mean = 7th term = sqrt(2*440*440) = 440*sqrt(2)

Answer A
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Re: A certain musical scale has has 13 notes, each having a [#permalink] New post 12 Jul 2013, 21:40
Pushpinder wrote:
Lowest frequency = 440
Highest frequency = 880
Lowest frequency (n) ^12 = Highest frequency
N^12 = 2 ---------------- 1
7th note = Lowest Frequency x (n)^6
7th note = 440 x (2)^6/12

Hence the answer is A


Pushpinder Ji I couldn't understand from here. Can u tell me please

Last edited by Bunuel on 12 Jul 2013, 23:28, edited 2 times in total.
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Re: A certain musical scale has has 13 notes, each having a [#permalink] New post 12 Jul 2013, 23:38
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dasikasuneel wrote:
Pushpinder wrote:
A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)

Lowest frequency = 440
Highest frequency = 880
Lowest frequency (n) ^12 = Highest frequency
N^12 = 2 ---------------- 1
7th note = Lowest Frequency x (n)^6
7th note = 440 x (2)^6/12

Hence the answer is A


Pushpinder Ji I couldn't understand from here. Can u tell me please


1st = 440
2nd = 440k
3rd = 440k^2
...
7th = 440k^6
...
13th = 440k^{12}=2*440=880 --> 440k^{12}=880 --> k^{12}=2 --> k=\sqrt[12]{2}.

Thus, 7th = 440k^6=440(\sqrt[12]{2})^6=440\sqrt{2}.

Answer: A.

Hope it's clear.

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Re: A certain musical scale has has 13 notes, each having a [#permalink] New post 26 Jul 2013, 04:50
Bunuel wrote:
dasikasuneel wrote:
Pushpinder wrote:
A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)

Lowest frequency = 440
Highest frequency = 880
Lowest frequency (n) ^12 = Highest frequency
N^12 = 2 ---------------- 1
7th note = Lowest Frequency x (n)^6
7th note = 440 x (2)^6/12

Hence the answer is A


Pushpinder Ji I couldn't understand from here. Can u tell me please


1st = 440
2nd = 440k
3rd = 440k^2
...
7th = 440k^6
...
13th = 440k^{12}=2*440=880 --> 440k^{12}=880 --> k^{12}=2 --> k=\sqrt[12]{2}.

Thus, 7th = 440k^6=440(\sqrt[12]{2})^6=440\sqrt{2}.

Answer: A.

Hope it's clear.



Hi Bunuel ,

It says that the ratio of ratio of a frequency to the next higher frequency is a fixed constant.
Doesnt that mean f1/f2 = k ??

Just a little lost.

Cheers
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Re: A certain musical scale has has 13 notes, each having a [#permalink] New post 26 Jul 2013, 05:40
Expert's post
heirapparent wrote:
Bunuel wrote:
dasikasuneel wrote:
Pushpinder Ji I couldn't understand from here. Can u tell me please


1st = 440
2nd = 440k
3rd = 440k^2
...
7th = 440k^6
...
13th = 440k^{12}=2*440=880 --> 440k^{12}=880 --> k^{12}=2 --> k=\sqrt[12]{2}.

Thus, 7th = 440k^6=440(\sqrt[12]{2})^6=440\sqrt{2}.

Answer: A.

Hope it's clear.



Hi Bunuel ,

It says that the ratio of ratio of a frequency to the next higher frequency is a fixed constant.
Doesnt that mean f1/f2 = k ??

Just a little lost.

Cheers
HeirApparent.


Does not matter how you write: f1/f2=constant --> f2/f1=1/constant.

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Re: A certain musical scale has has 13 notes, each having a [#permalink] New post 26 Jul 2013, 07:27
Hi Bunuel ,

It says that the ratio of ratio of a frequency to the next higher frequency is a fixed constant.
Doesnt that mean f1/f2 = k ??

Just a little lost.

Cheers
HeirApparent.[/quote]

Does not matter how you write: f1/f2=constant --> f2/f1=1/constant.[/quote]

Got it ...

Cheers

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Re: A certain musical scale has has 13 notes, each having a [#permalink] New post 24 Sep 2013, 19:12
In a geometric progression, the median is the geometric mean given by SQRT (First * Last).
Here, First is 440, Last is 2*440 = 880 and 7th Note is the median, so it's value = SQRT (440*880) = SQRT (440*440*2) = 440*SQRT(2)
A is correct.
Re: A certain musical scale has has 13 notes, each having a   [#permalink] 24 Sep 2013, 19:12
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